My Math Forum > Math Prove by mathematical induction

 Math General Math Forum - For general math related discussion and news

 March 26th, 2018, 10:10 AM #1 Newbie   Joined: Mar 2018 From: Trinidad Posts: 2 Thanks: 0 Prove by mathematical induction chelsea_lutchman@icloud.com Prove by Mathematical Induction n sigma r^3 = n^2(n+1)^2/4 r = 1 Can someone show me how the do these sums? I've asked my lecturer for help but all he tells me is that it's too easy and he shouldn't have to help me. It would have been easier if he could at least give me an example about how it's done but he didn't. Google hasn't been a big help either. I'm not seeing any examples of induction with sigma that I can understand. Last edited by notchelsea; March 26th, 2018 at 10:48 AM.
 March 26th, 2018, 11:50 AM #2 Newbie   Joined: Mar 2018 From: Trinidad Posts: 2 Thanks: 0 so far I have 1 sigma r^3 = 1^2(1+1)^2/2 r=1 1 = 1(4)/2 1 = 4/2 1 = 2 I'm not sure what to do after this for the k+1 case. (took too long to edit so this is what I wanted to add)
 March 26th, 2018, 12:40 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,030 Thanks: 420 Couple of things. It is frequently a good idea to do the first few numbers by hand. $\displaystyle \sum_{r=1}^1 r^3 = 1^3 = 1 = \dfrac{4}{4} = \dfrac{1 * 2^2}{4} = \dfrac{1^2(1 + 1)^2}{4}.$ Congratulations. You have already completed the first part of a proof by induction. $\displaystyle \sum_{r=1}^2 r^3 = 1^3 + 2^3 = 9 = \dfrac{36}{4} = \dfrac{4 * 9}{4} = \dfrac{2^2(2 + 1)^2}{4}.$ $\displaystyle \sum_{r=1}^3 r^3 = 1^3 + 2^3 + 3^3 = 36 = \dfrac{4 * 36}{4} = \dfrac{9 * 16}{4} = \dfrac{3^2(3 + 1)^2}{4}.$ Doing this let's you understand what the theorem means, gives some confidence that it may well be true, and sometimes gives you a hint at a proof. OK. The proof is in two stages. $\text {PROVE: } \displaystyle n \in \mathbb N^+ \implies \sum_{r=1}^n r^3 = \dfrac{n^2(n + 1)^2}{4}.$ $n = 1 \implies \displaystyle \sum_{r=1}^n r^3 = 1^3 = 1 = \dfrac{1^2(1 + 1)^2}{4}.$ First stage is done. $\therefore \exists \ k \in \mathbb N^+ \text { such that } k \ge 1 \text { and } \displaystyle \sum_{r= 1}^k r^3 = \dfrac{k^2(k + 1)^2}{4}.$ Now what you FREQUENTLY do is take the left hand side of the equation for k + 1 and try to break it down into the known expression in k and a simple expression in k + 1. $\displaystyle \sum_{r=1}^{k+1}r^3 =$ $\displaystyle \left ( \sum_{r=1}^k r^3 \right ) + (k + 1)^3 =$ $\dfrac{k^2(k + 1)^2}{4} + k^3 + 3k^2 + 3k + 1 =$ $\dfrac{k^4 + 2k^3 + k^2}{4} + \dfrac{4k^3 + 12k^2 + 12k + 4}{4} =$ $\dfrac{k^4 + 6k^3 + 13k^2 + 12k + 4}{4} =$ $\dfrac{(k^4 + 4k^3 + 4k^2) + (2k^3 + 8k^2 + 8k) + (k^2 + 4k + 4)}{4} =$ $\dfrac{k^2(k^2 + 4k + 4) + 2k(k^2 + 4k + 4) + 1(k^2 + 4k + 4)}{4} =$ $\dfrac{(k^2 + 2k + 1)(k^2 + 4k + 4)}{4} = \dfrac{(k + 1)^2(k + 2)^2}{4} =$ $\dfrac{(k + 1)^2\{(k + 1) + 1\}^2}{4}.$ Done. Last edited by JeffM1; March 26th, 2018 at 12:51 PM. Reason: Fixed missing parenthesis
March 26th, 2018, 12:49 PM   #4
Senior Member

Joined: May 2016
From: USA

Posts: 1,030
Thanks: 420

Quote:
 Originally Posted by notchelsea so far I have 1 sigma r^3 = 1^2(1+1)^2/2 r=1 1 = 1(4)/2 1 = 4/2 1 = 2 I'm not sure what to do after this for the k+1 case. (took too long to edit so this is what I wanted to add)
Do you see that you messed up the case for n = 1. When you get an answer that
1 = 2, you can be fairly sure something came off the rails.

 Tags induction, mathematical, prove

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post DecoratorFawn82 Algebra 4 October 7th, 2017 03:26 AM numeriprimi Elementary Math 2 September 30th, 2017 11:28 AM mitch08 Computer Science 1 October 4th, 2015 08:39 PM wannabemathlete Algebra 4 October 23rd, 2014 10:01 PM Phatossi Algebra 4 November 10th, 2012 05:27 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top